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owlpride
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Homework Statement
Assume that f(z) is analytic and that f'(z) is continuous in a region that contains a closed curve [tex]\gamma[/tex]. Show that
[tex] \int_\gamma \overline{f(z)} f'(z) dz[/tex]
is purely imaginary.
Homework Equations
If f(z) is holomorphic on the region containing a closed curve [tex]\gamma[/tex] or if f(z) has a primitive (we have not yet established a relationship between these two properties), then
[tex]\int_\gamma f(z)dz = 0.[/tex]
And while we did not prove that analytic functions are holomorphic, that's not hard to verify if necessary.
The Attempt at a Solution
If we let [tex]f(z) = u(z) + i v(z)[/tex], where u and v are functions from the complex plane to the real line, then
[tex]\int_\gamma \overline{f(z)} f'(z) dz = \int f(z) f'(z) dz - 2 i \int v(z) f'(z) dz = 0 - 2 i \int_\gamma v(z) f'(z)dz[/tex]
So I have to verify that the latter integral is real valued, and this is where I am stuck. I could parametrize [tex]z = \gamma(t)[/tex] where [tex]t \in [0,1][/tex] but I am not sure that helps me in any way.
[tex]- 2 i \int_\gamma v(z) f'(z)dz = -2 i \int_0^1 v(\gamma(t)) f'(\gamma(t)) \gamma'(t) dt = 2 i \int_0^1 v'(\gamma(t))f'(\gamma(t))\gamma'(t) \gamma'(t) dt[/tex]
(The last step comes from integration by parts, with one term = 0 because gamma is a closed curve.)
Is there any reason to believe that this last integral is real-valued?
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